Abstract

In crystal structures, if each crystal cell is a special rectangular box (length × width × height=23 × 11.5 × 11.5 angstroms3), then how to pack these cells onto a enough hard base of a (46 × 46 angstroms2) square in order to make the crystal reach the stablest/optimal structure as high as the crystal it is? This is a practical problem that we are always thinking in our mind. This short article is to present an algorithm to solve this problem.

Keywords

Step 1: Put 2 boxes on one of the four boundaries (denoted as B1) of the base, with their lengths occupying B1.

Step 2: Put 1 box on each of the neighbored boundaries (denoted as B2 and B3) respectively, with each length side occupying each boundary respectively.

Step 3: Put 2 boxes on the remaining boundary (denoted as B4) of the base, with their lengths occupying B4.

Step 4: Now a 23 × 23 angstroms2 square is inner (or in the center of) the base. Put 2 boxes onto the inner square, with their length sides paralleling to B1 and/or B4. Now we finished the layout of the 1st layer.

Step 5: Determine how many layers that you will lay the boxes, according to your own constraints. Once the number of layers was determined, we will repeat Step 6 to finish the layout onto the inner 23 × 23 angstroms2 square.

Step 6: Put 2 boxes in the even number of the layer, with their length sides paralleling to B2 and/or B3; put 2 boxes in the odd number of the layer, with their length sides paralleling to B1 and/or B4.

Step 7: Now we lay boxes onto the outside of the inner 23 × 23 angstroms2 square: repeating Steps 8 ~ 9 until reaching the maximum number of layers.

Step 8: In the even number of the layer, put 1 box on both sides paralleling to B1 and B4 respectively; in the odd number of the layer, put 1 box on both sides paralleling to B2 and B3 respectively.

Step 9: In the even number of the layer, put 2 boxes on both sides paralleling to B2 and B3 respectively; in the odd number of the layer, put 2 boxes on both sides paralleling to B1 and B4 respectively.

The above algorithm is illuminated in Figure 1 in a workshop. If the maximum number of layers goes to infinity, and each layer produced by the above algorithm is looked as one solution of the algorithm, then the above algorithm is convergent in the strict algorithm convergence theory [1-3].

Figure 1: An illumination of the Algorithm.

Acknowledgements

This research was supported by a Victorian Life Sciences Computation Initiative (VLSCI) grant numbered VR0063 on its Peak Computing Facility at the University of Melbourne, an initiative of the Victorian Government, Australia.

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Review summary

Fahida Posted on
Sep 29 2016 at 6:57 pm
This is short article describes the how to pack the rectangular-box cells optimally for the crystal structure. The author describes an algorithm to pack these cells onto a enough hard base of a (46 × 46 angstroms2) square in order to make the crystal reach the stablest/optimal structure as high as the crystal it is. This manuscript is well written. Given algorithm may help the crystal cell reach stablest/optimal structure.